Answer:
P2 = 1125 mmHg
Explanation:
Gas Pressure Calculation
To solve this problem, we need to use the combined gas law, which relates the pressure, volume, and temperature of a gas under different conditions. The combined gas law is given by:
(P1V1)/T1 = (P2V2)/T2
where P1, V1, and T1 are the initial pressure, volume, and temperature of the gas, and P2, V2, and T2 are the final pressure, volume, and temperature.
Let's start by calculating the initial conditions:
P1 = 750 mmHg
V1 = 120 cm^3
T1 = 25°C + 273.15 = 298.15 K (temperature in Kelvin)
Now we can plug in these values and solve for P2:
(P1V1)/T1 = (P2V2)/T2
(750 mmHg x 120 cm^3) / 298.15 K = (P2 x 150 cm^3) / (40°C + 273.15)
Simplifying this equation, we get:
P2 = (750 mmHg x 120 cm^3 x (40°C + 273.15)) / (298.15 K x 150 cm^3)
P2 = 1125 mmHg
Therefore, the pressure of the gas would increase to 1125 mmHg if its volume increased to 150 cm^3 at 40°C.
ChatGPT
Among the elements of the main group, the first ionization energy increases
from left to right across a period.
from right to left across a period.
when the atomic radius increases.
down a group.
The first ionisation energy increases over time from left to right among the major group of elements. answer is option (a).
What is Ioniztion?When an element loses its valence electron, its oxidation number increases (a process known as oxidation), and this energy loss is known as ionisation (Ei).
Earth alkaline metals, which are located immediately next to alkaline metals, have higher ionisation energies than alkaline metals because they have two valence electrons, while alkaline metals, which are located far left in the main group, have the lowest ionisation energies and are easiest to remove.
Because they contain a large number of valence electrons, nonmetals are far to the right in the main group and have the highest ionisation energy.
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The complete question is,
Among the elements of the main group, the first ionization energy increases
a. from left to right across a period.
b. from right to left across a period.
c. when the atomic radius increases.
d. down a group.
A student adds 5.48 g of vitamin C ( ascorbic acid, C6H8O6 ) to 50.0 mL of water. What is the molarity of the situation
The molarity of the ascorbic acid ([tex]C_6H_8O_6[/tex]) solution is 0.622 M.
To find the molarity of solution, we need to first calculate number of moles of ascorbic acid present in solution, using the formula:
moles = mass / molar mass
The molar mass of ascorbic acid is:
6(12.01 g/mol) + 8(1.01 g/mol) + 6(16.00 g/mol) = 176.12 g/mol
So, the number of moles of ascorbic acid present in the solution is:
moles = 5.48 g / 176.12 g/mol = 0.0311 mol
Next, we need to calculate the volume of the solution in liters, using the conversion:
[tex]1 mL = 1 * 10^{-3} L[/tex]
50.0 mL x 1 L / 1000 mL = 0.0500 L
Finally, we can calculate the molarity of the solution:
molarity = 0.0311 mol / 0.0500 L = 0.622 M
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100 mL of 3 M HCl can be neutralized with exactly 200 mL of 1.5 M NaOH. If this reaction is done in a coffee cup calorimeter, what temperature change would you expect to observe?
100 mL of 3 M HCl can be neutralized with exactly 200 mL of 1.5 M NaOH. If this reaction is done in a coffee cup calorimeter, temperature would increase.
A chemical reaction known as neutralisation occurs when an acid and a base quantitatively react with one another. Alternate spellings include Neutralisation. The pH in the neutralised solution is determined by the acidity of the reactant.
Have you ever overindulged in spicy food and felt your stomach start to burn? This results from the stomach producing acid. The use of an antacid, which counteracts the effects of acid, can solve this issue; this process is known as a neutralisation response. 100 mL of 3 M HCl can be neutralized with exactly 200 mL of 1.5 M NaOH. If this reaction is done in a coffee cup calorimeter, temperature would increase.
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climate on river and streams
Due to changes in stream temperature, stream flow due to shortages or increased storms, and other stressors which can affect ecosystem health, these aquatic ecosystems are at risk from climate change.
The long-term trend of the weather in a place is called the climate. Hour by hour, day by day, month by month, or even year by year, the weather might change. Due to changes in stream temperature (which results in a corresponding drop in oxygen levels).
Stream flow due to shortages or increased storms, and other stressors (such as increased storm runoff such as nutrients, pollutants, along with sediment) which can affect ecosystem health, these aquatic ecosystems are at risk from climate change.
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PLEASE HELP!!!!!!!!!!
A teaspoon of salt, NaCl has a mass of about 5.0 g. How many formula units are in a teaspoon of salt?
SHOW WORK PLS!!!!
The molar mass of NaCl is 58.44 g/mol, which means that one mole of NaCl contains 6.022 x 10^23 formula units (Avogadro's number).
To determine the number of formula units in a teaspoon of salt, we need to first determine how many moles of NaCl are present in 5.0 g of salt. This can be done using the following formula:
moles = mass / molar mass
moles = 5.0 g / 58.44 g/mol = 0.0854 mol
Next, we can use Avogadro's number to convert moles of NaCl to formula units:
formula units = moles x Avogadro's number
formula units = 0.0854 mol x 6.022 x 10^23 formula units/mol = 5.14 x 10^22 formula units
Therefore, there are approximately 5.14 x 10^22 formula units of NaCl in a teaspoon of salt.
Fifteen 15kg of iron lll oxide was used in a reaction to produce iron.calculate the mass of iron produced in the reaction
Answer: 10.51 kg of iron is produced in the reaction.
Explanation: The balanced equation for the reaction between iron III oxide and iron is:
2Fe2O3 + 3C → 4Fe + 3CO2
From the equation, we can see that 2 moles of Fe2O3 react to produce 4 moles of Fe. We can use the molar mass of Fe2O3 to convert the given mass to moles, and then use the mole ratio to calculate the moles of Fe produced. Finally, we can convert the moles of Fe to mass using the molar mass of Fe.
The molar mass of Fe2O3 is:
2(55.85 g/mol Fe) + 3(16.00 g/mol O) = 159.69 g/mol Fe2O3
So, 15 kg (or 15000 g) of Fe2O3 is equal to:
15000 g / 159.69 g/mol = 94.03 mol Fe2O3
According to the balanced equation, 2 moles of Fe2O3 produce 4 moles of Fe. So, 94.03 moles of Fe2O3 will produce:
4/2 x 94.03 = 188.06 moles of Fe
The molar mass of Fe is 55.85 g/mol, so the mass of Fe produced is:
188.06 mol x 55.85 g/mol = 10507.8 g or 10.51 kg
1) The last state of matter we will study is gases. Gas quantities are measured using four common
variables:
a) P for ____which is often measured in units of ____, ____, ____, and ____
b) V for___ which is often measured in units of___,___,___, and ___
c) T for _____ _____ which must be in units ____.
d) n for ____, which is often found by converting from grams of a gas.
1) The last state of matter we will study is gases.
a) P for Pressure which is often measured in units of atmospheres (atm), milliamps (mm Hg), pounds per square inch (psi), pascals (Pa)
b) V for Volume which is often measured in units of liters (L), cubic meters (m3), cubic feet (ft3), cubic inches (in3)
c) T for Temperature which must be in units Kelvin (K) or Celsius (°C).
d) n for Number of moles , which is often found by converting from grams of a gas.
5.0g of hydrochloric acid is mixed with 24.0g of magnesium hydroxide.
what mass of water is produced?
From the balanced chemical equation we can see tha the mass of water produced is 2.47 grams.
What mass of water is produced?The balanced chemical equation for the reaction between hydrochloric acid (HCl) and magnesium hydroxide (Mg(OH)2) is:
2 HCl + Mg(OH)2 → MgCl2 + 2 H2O
From the equation, we can see that 2 moles of hydrochloric acid react with 1 mole of magnesium hydroxide to produce 2 moles of water. To calculate the mass of water produced, we need to determine the limiting reactant, which is the reactant that is completely consumed first and determines the amount of product formed.
First, let's calculate the number of moles of each reactant:
Mass of HCl = 5.0 g
Molar mass of HCl = 36.46 g/mol
Number of moles of HCl = 5.0 g / 36.46 g/mol = 0.137 mol
Mass of Mg(OH)2 = 24.0 g
Molar mass of Mg(OH)2 = 58.33 g/mol
Number of moles of Mg(OH)2 = 24.0 g / 58.33 g/mol = 0.411 mol
According to the balanced chemical equation, 2 moles of HCl react with 1 mole of Mg(OH)2 to produce 2 moles of water. Therefore, the stoichiometry of the reaction requires 2 moles of HCl for every 1 mole of Mg(OH)2. Since we have only 0.137 mol of HCl and 0.411 mol of Mg(OH)2, the HCl is the limiting reactant because it is completely consumed first.
The molar ratio of HCl to water is 2:2, which simplifies to 1:1. Therefore, the number of moles of water produced is also 0.137 mol.
Now, let's calculate the mass of water produced using the molar mass of water:
Molar mass of water (H2O) = 18.02 g/mol
Mass of water produced = 0.137 mol x 18.02 g/mol = 2.47 g
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CAN SOMEONE HELP WITH THIS QUESTION?
FROM Spectrophotometric Determination of Cobalt (ll)
Answer:
If a solution appears red, it is most likely that green light is being absorbed the strongest. When light passes through a solution, the solution absorbs certain colors of light while allowing others to pass through. The color that we see is the color that is not absorbed and is transmitted through the solution. In the case of a red solution, red light is transmitted while green light is absorbed. This selective absorption of light is due to the specific chemical composition of the solution.
Calculate the cell potential for the following reaction that takes place in an electrochemical cell at 25°C.
Cu(s) | Cu2+(aq, 0.0032 M) || Cu2+(aq, 4.48 M) | Cu(s)
Calculate the cell potential for the following reaction that takes place in an electrochemical cell at 25°C.
Cu(s) | Cu2+(aq, 0.0032 M) || Cu2+(aq, 4.48 M) | Cu(s)
+0.186 V
0.00 V
+0.093 V
+0.34 V
+0.052 V
Okay, let's solve this step-by-step:
1) The standard reduction potentials for Cu2+/Cu are: E°Cu2+/Cu = +0.34 V.
This is the reduction potential when [Cu2+] = 1 M and [Cu] = 1 M.
2) The actual reduction potential (Ered) depends on the concentrations of oxidized and reduced species.
Here,
[Cu2+] = 4.48 M on the right side.
[Cu2+] = 0.0032 M on the left side.
3) Ered = E° + 0.0591 log([oxidized]/[reduced]) (Nernst equation)
So for the right side:
Ered = +0.34 + 0.0591 log(4.48/1) = +0.34 + 0.186 = +0.526 V
And for the left side:
Ered = +0.34 + 0.0591 log(0.0032/1) = +0.34 - 0.093 = +0.247 V
4) The cell potential (Ecell) is the difference between the two half-cell potentials:
Ecell = +0.526 - 0.247 = +0.279 V
So the cell potential for the given reaction at 25°C is +0.279 V.
Let me know if you have any other questions!
(ii) 10% of the atoms in a sample of element E have a mass number of 22. All the other atoms in this sample have a mass number of 20. Calculate the relative atomic mass of element E.
Answer:
Explanation:
The total mark for this paper is 60.
t The marks for each question are shown in brackets
– use this as a guide as to how much time to spend on each question.
t Questions labelled with an asterisk (*) are ones where the quality of your
written communication will be assessed
– you should take particular care with your spelling, punctuation and grammar, as
well as the clarity of expression, on these questions.
Advice
t Read each question carefully before you start to answer it.
t Keep an eye on the time.
t Try to answer every question.
t Check your answers if you have time at the end.
Calculate the cell potential (Ecell) at 25oC (298 K) for the following reaction if the Cu2+ ion concentration is 0.064 M and the Fe2+ ion concentration is 0.645 M.
Fe(s) + Cu2+(aq) → Fe2+(aq) + Cu(s)
Half-reaction Standard Reduction Potential (V)
Fe2+(aq) + 2e−→ Fe(s) −0.440
Cu2+(aq) + 2e−→ Cu(s) +0.337
R = 8.31 V/mol·K
F = 96500 C/mol
The cell potential (Ecell) at 25°C (298 K) for the given reaction is 1.065 V which means that the reaction is spontaneous because the calculated Ecell is positive
The cell potential (Ecell) for the given reaction can be calculated using the Nernst equation:
Ecell = E°cell - (RT ÷ nF) × ln(Q)
where E°cell = standard cell potential, R = gas constant, T = temperature in Kelvin, n = number of electrons transferred in the reaction, F = Faraday constant, and Q is the reaction quotient.
Since two electrons are transported in the half-reactions, n in this instance equals 2. With the help of the species concentrations, it is possible to determine the reaction quotient Q:
Q = [Fe2+] ÷ [Cu2+]
Q = 0.645 ÷ 0.064
Q = 10.078
Now, we can calculate the Ecell:
Ecell = E°cell - (RT ÷ nF) × ln(Q)
Ecell = (0.337 - (-0.440)) - (8.31 × 298 ÷ (2 × 96500)) × ln(10.078)
Ecell = 1.065 V
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How many grams of PbBr2 will precipitate when excess FeBr2 solution is added to 58.9 mL of 0.505 M Pb(NO3)2 solution?
when a sufficient [tex]FeBr_{2[/tex] solution is introduced to 58.9 mL of 0.505 M [tex]Pb(NO_3)_2[/tex] solution, 5.45 grammes of [tex]PbBr_2[/tex] will precipitate.
Calculation-We can figure out the amount of lead(II) bromide (PbBr2) that will precipitate by using the stoichiometry of the reaction between lead(II) nitrate [tex]Pb(NO_3)_2[/tex] and iron(II) bromide [tex](FeBr_2).[/tex]
The reaction's chemically balanced equation is as follows:
[tex]PbBr_2(s) + 2 Fe(NO_2)2 = Pb(NO_2)2(aq) + 2 FeBr_2 (aq)(aq)[/tex]
According to the equation, two moles [tex]FeBr_2[/tex] react with one mole of [tex]Pb(NO_3)_2[/tex] to create one mole [tex]FeBr_2[/tex]. The amount [tex]PbBr_2[/tex] that will precipitate will therefore be equal to half of the amount of [tex]Pb(NO_3)_2[/tex] that is present in the solution.
We can use the following formula to get the amount of[tex]Pb(NO_3)_2[/tex] in moles:
moles = volume x concentration
where volume is given as 58.9 mL, which we will divide by 1000 to get litres, and concentration is given as 0.505 M:
[tex]moles of Pb(NO3)2 = 0.505 M x 0.0589 L = 0.0297 moles[/tex]
Therefore, the amount of PbBr2 that will precipitate is:
[tex]PbBr2 moles =0.0297 moles / 2 moles, or 0.01485 moles.[/tex]
Finally, using the molar mass of [tex]PbBr_2[/tex], which is 367.01 g/mol, we can convert the number of moles [tex]PbBr_2[/tex] to grammes:
[tex]mass of PbBr2 = 0.01485 moles x 367.01 g/mol = 5.45 g[/tex]
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see image attached please and thank you
hope it helps:)
[tex]a. \: Ca + Cl _{2} → CaCl _{2} \\
\\ b. \: Cl _{2}+H _{2} O+ NaOH → \\ NaCl+ H _{2}O \\ \\
c. \: \: \: H _{2} SO _{4} +CaCO _{3} → \\ CaSO _{4} +H _{2}O+CO _{2} \\ \\
d. \: \: Fe+Cu(NO _{3}) _{2} → \\ Fe(NO _{3} ) _{2}+Cu[/tex]
brainliest pls ｡◕‿◕｡
Hydrogen Bonds: A specific type of dipole-dipole attraction results from the interaction of a hydrogen (H) atom and a weak electronegative atom
true or false
True. A specific type of dipole-dipole attraction results from the interaction of a hydrogen (H) atom and a weak electronegative atom
Why is the above statement true?A particular kind of dipole-dipole interaction known as a hydrogen bond occurs when a hydrogen atom is covalently connected to an element that is strongly electronegative, such as nitrogen (N), oxygen (O), or fluorine (F), and another weakly electronegative atom that is present nearby. The strongly electronegative atom draws the electrons in the hydrogen-self bond, leaving the hydrogen with a partial positive charge and the electronegative atom with a partial negative charge. As a result, this interaction takes place. A hydrogen bond, a relatively potent attraction, is produced when the hydrogen atom's partial positive charge interacts electrostatically with the surrounding electronegative atom's partial negative charge. Many biological processes, such as the binding of DNA base pairs, protein folding, and cell division, depend on hydrogen bonding.
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Convert 231 μm3 to cm3 .
To convert micrometers cubed (μm^3) to cubic centimeters (cm^3), we need to divide by 10^12 (since there are 10^12 μm^3 in 1 cm^3).
So, to convert 231 μm^3 to cm^3, we can use the following formula:
231 μm^3 ÷ (10^12 μm^3/cm^3) = 2.31 x 10^-10 cm^3
Therefore, 231 μm^3 is equal to 2.31 x 10^-10 cm^3.
:)
To convert micrometers cubed (μm^3) to cubic centimeters (cm^3), we need to divide by 10^12 (since there are 10^12 μm^3 in 1 cm^3).
So, to convert 231 μm^3 to cm^3, we can use the following formula:
231 μm^3 ÷ (10^12 μm^3/cm^3) = 2.31 x 10^-10 cm^3
Therefore, 231 μm^3 is equal to 2.31 x 10^-10 cm^3.
:)
Locating the epicenter of an earthquake lab
Eruption triangulation is a method for locating an earthquake's epicenter.
What is epicenter?The area right above the spot in the Earth's crust where an earthquake originates or begins is known as the epicenter. It marks the spot on the Earth's surface where the earthquake's seismic waves first touchdown.
The distance between the earthquake's epicenter and a number of seismograph stations is calculated using seismograms, which are records of the ground motion brought on by an earthquake.
The following are the steps to find an earthquake's epicenter:
A minimum of three separate seismograph stations should have data collected. The epicenter's distance from each station is determined by keeping track of the time the earthquake waves arrived at each station.
Map out the positions of each seismograph station.
Draw circles with an equal radius around each seismograph station using the distance information. Each circle's radius is equal to how far away the station is from the epicenter.
At the place where the circles converge, there is an epicenter.
It is crucial to keep in mind that determining the epicenter of an earthquake is not an exact science, and the precision of the position will vary depending on a number of variables, such as the caliber of the seismograms and the distance between the earthquake and the seismograph stations.
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What is the only part of the comet that exists when it is further than 5 A.U from the sun?
Sublimation and the coma is the only part of the comet that exists when it is further than 5 A.U from the sun.
Generally by the time that the comet comes within about 5 AU of the Sun, sublimation has generally formed a noticeable atmosphere that can easily escape the comet's weak gravity. The coma basically forms as the escaping atmosphere which drags away dust particles that have been mixed with the sublimating ice.
Generally when a comet is at a great distance away from the Sun, it exists as a dirty snowball several kilmometers across. But when it comes closer to the Sun, the warming of its surface causes its materials to melt and vaporize which produces the comet's characteristic tail.
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What is molecular weight of a substance given that 1.22g of the sample was vaporised in 100ml flask at 45°C and 687mmHg.
To calculate the molecular weight of a substance from its vapor density, we can use the following formula:
Molecular weight = (RT/P) x d
where:
R is the gas constant (0.0821 L atm/mol K)
T is the temperature in Kelvin
P is the pressure in atm
d is the vapor density (in g/L) of the substance
First, we need to calculate the vapor density of the substance using the given information. We can use the ideal gas law to find the number of moles of the substance in the flask:
PV = nRT
where:
P is the pressure (687 mmHg = 0.903 atm)
V is the volume (100 mL = 0.1 L)
n is the number of moles of gas
R is the gas constant (0.0821 L atm/mol K)
T is the temperature in Kelvin (45°C = 318 K)
Solving for n, we get:
n = PV/RT = (0.903 atm)(0.1 L)/(0.0821 L atm/mol K)(318 K) = 0.00372 mol
Next, we can use the mass and volume information to find the density of the substance:
density = mass/volume = 1.22 g/0.1 L = 12.2 g/L
Since the vapor density is half of the density of the substance in the liquid state, we can calculate the vapor density:
vapor density = density/2 = 6.1 g/L
Finally, we can use the formula above to find the molecular weight:
Molecular weight = (RT/P) x d
Molecular weight = (0.0821 L atm/mol K)(318 K)/(0.903 atm) x 6.1 g/L
Molecular weight = 92.2 g/mol
Therefore, the molecular weight of the substance is 92.2 g/mol.
The cell diagram for the lead-acid cell that is used in automobile and truck batteries is
Pb(s)∣∣PbSO4(s)∣∣H2SO4(aq)∣∣PbO2(s),PbSO4(s)∣∣Pb(s)
The comma between PbO2(s) and PbSO4(s) denotes a heterogeneous mixture of the two solids. The right-hand lead electrode is nonreactive.
Write the balanced equation for the net cell reaction.
Look up standard potentials for the oxidation and reduction half-reactions, and then calculate the value of ∘cell
Calculate the value of Δ∘rxn
Calculate the value of cell at 25 ∘C if [H2SO4]=10.0 M
How many lead-acid cells are in a 12 Vcar battery? Round to the nearest integer.
number of lead-acid cells:
The net cell reaction is described by the equation: Pb(s) + PbO2(s) + 4H2SO4(aq) 2PbSO4(s) + 2H2O(l). 2.14 V and -0.36 V, respectively, are the standard potentials for the oxidation and reduction half-reactions. The result is that cell = 2.50 V.
The computed value of rxn is 2.50 V, which is the difference between the two standard potentials. The computed cell potential with [H2SO4] = 10.0 M at 25 C is 2.50 V. Six lead-acid cells connected in series make up a 12 V automobile battery.
Six cells are required to obtain 12 V because each cell has a voltage of 2.0 V. Consequently, there are six lead-acid cells in a 12 V automobile battery.
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Complete the balanced molecular chemical equation for the reaction below. If no reaction occurs, write NR after the reaction arrow. Be sure to include the proper phases for all species within the reaction.
CsCl(aq) + K3PO4(aq) —>
The balanced equation for the given chemical reaction can be written as 3CsCl + K[tex]_3[/tex]PO[tex]_4[/tex]→ Cs[tex]_3[/tex] (PO[tex]_4[/tex]) + 3KCl.
An equation per a chemical reaction is said to be balanced if both the reactants plus the products have the same number of atoms and total charge for each component of the reaction. In other words, each component of the reaction have an equal balance of mass and charge.
The components and outcomes of a chemical reaction are listed in an imbalanced chemical equation, but the amounts necessary to meet the conservation of mass are not specified. The balanced equation for the given chemical reaction can be written as 3CsCl + K[tex]_3[/tex]PO[tex]_4[/tex]→ Cs[tex]_3[/tex] (PO[tex]_4[/tex]) + 3KCl.
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2LiO
1. Determine the type of compound
2. Name the compound
3. Determine the chemical formula for the compound
Answer:
1. Ionic Compound
2.Lithium Oxide
3. Li2O
Explanation:
This is an Ionic Compound because (Lithium) the metal or cation is giving away electrons to the oxygen a nonmental or anion so both can have a full valence electron shell or have all the electrons on the outer most shell.
This is also Lithium Oxide because the metal goes first then the nonmetal and the ending of the non metal is ending with an ide.
The chemical formula is the periodic table symbols (usually the metal first) and then the charges are switched. E.g Sample A has charge of 3 and sample B has a charge of 2. The formula would be A2B3. If the charge is 1 then you can leave it blank because it is understood. E.g is sample B had a charge of 1 then it would be AB3
Suppose that a certain quantity of methane occupies a volume of 0.138 L under a pressure of 300 atm at 200 °C, and the volume required at 600 atm at 0 °C. For 300 atm and at 200 °C, Z=1.067, while for 600 atm at 0 °C, Z=1.367.
Answer:
Therefore, the volume required at 600 atm and 0 °C is 0.319 L.
Explanation:
We can use the ideal gas law to solve for the number of moles of methane present, assuming ideal gas behavior at both conditions:
PV = nRT
At 300 atm and 200 °C:
n = PV/RT = (300 atm * 0.138 L) / [(0.08206 L atm mol^-1 K^-1) * (200 + 273.15) K * 1.067]
n = 2.451 mol
At 600 atm and 0 °C:
n = PV/RT = (600 atm * V2) / [(0.08206 L atm mol^-1 K^-1) * (273.15 K) * 1.367]
n = 7.682 V2
Since the number of moles of methane must be the same at both conditions:
2.451 mol = 7.682 V2
Solving for V2:
V2 = 0.319 L
Therefore, the volume required at 600 atm and 0 °C is 0.319 L.
A flexible container at an initial volume of 3.10 L contains 3.51 mol of gas. More gas is then added to the container until it reaches a final volume of 17.1 L. Assuming the pressure and temperature of the gas remain constant, calculate the number of moles of gas added to the container.
The container received 15.7 moles of gas.
What is volume?Volume is a unit used to describe how much space an object or substance occupies. It is a physical quantity that, depending on the situation, is typically measured in measures like liters, cubic meters, or cubic feet.
How do you determine it?The ideal gas law can be used since the gas's temperature and pressure are constant:
PV = nRT
n is the number of moles, P is the pressure, V is the volume, R is the gas constant, and T is the temperature.
The following equation can be used to link the initial and final volumes because the pressure and temperature are constant.
V1 / n1 = V2 / n2
where V1 and n1 represent the beginning volume and molecular count, and V2 and n2 represent the ultimate volume and molecular count.
To solve for n2 we can rearrange this equation as follows:
n2 = (V2 / V1) * n1
Plugging in the values we are familiar with yields:
n2 = (17.1 L / 3.10 L) * 3.51 mol.
n2 = 19.2 mol
As a result, the container was filled with the following amount of gas:
n2 - n1= 19.2 mol - 3.51 mol = 15.7 mol
As a result, the container received 15.7 moles of gas.
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A sample of air from a factory smokestack measured at 35 °C contained SO₃ at a partial pressure of 8.75 torr. What mass, in g, of SO₃ is in 1.00 L of the air sample?
The mass, in grams, of SO₃ in 1.00 L of the air sample is 27.48 g.
To determine the mass of SO₃ in 1.00 L of an air sample, we need to use the ideal gas law:
PV = nRT
where P is the partial pressure of SO₃, V is the volume of the sample (1.00 L), n is the number of moles of SO₃, R is the gas constant, and T is the temperature in Kelvin (308 K)
n = PV/RT
n = (8.75 torr) × (1.00 L) ÷ (0.0821 L·atm/mol·K) × (308 K)
n = 0.343 mol
To convert this to mass, we need to use the molar mass of SO₃, which is 80.06 g/mol. Therefore, the mass of SO₃ in 1.00 L of the air sample is:
mass = n × molar mass
mass = 0.343 mol × 80.06 g/mol
mass = 27.48 g
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what is the advantages of gold
Answer:
Gold is the most malleable and ductile metal. It is soft and usually alloyed to give it more stability as it is easily bent. It is a great conductor of heat and electricity, but its greatest strength comes from the fact that does not react with oxygen. It is found that gold is unaffected by air, water, bases and most acids.
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The absorbance of this solution at this wavelength would be 0.287.
How to find the absorbance ?We can use the relationship between percent transmittance (%T) and absorbance (A) :
% T = 10 ^ ( - A )
Rearranging this equation, we can solve for A:
A = - log (%T / 100 )
Substituting the given value, we get:
A = - log ( 51. 6 / 100) = - log (0. 516) = 0. 287
Therefore, the absorbance of this solution at 550 nm is 0.287 .
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Calculate the mass of butane needed to produce 70.8 g of carbon dioxide
The mass of butane, C₄H₁₀ needed to produce 70.8 g of carbon dioxide, CO₂ is 23.3 g
How do i determine the mass of butane needed?First, we shall write the balanced equation for the reaction. Details below:
2C₄H₁₀ + 13O₂ -> 8CO₂ + 10H₂O
Molar mass of C₄H₁₀ = 58 g/molMass of C₄H₁₀ from the balanced equation = 2 × 58 = 116 g Molar mass of CO₂ = 44 g/molMass of CO₂ from the balanced equation = 8 × 44 = 352 gFrom the balanced equation above,
352 g of CO₂ were obtained from 1 16 g of C₄H₁₀
Finally, we shall determine the mass of butane, C₄H₁₀ neeeded to produce 70.8 g of carbon dioxide, CO₂. Details below:
From the balanced equation above,
352 g of CO₂ were obtained from 116 g of C₄H₁₀
Therefore,
70.8 g of CO₂ will be obtain from = (70.8 × 116) / 352 = 23.3 g of C₄H₁₀
Thus, the mass of butane needed is 23.3 g
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Provide feedback to me if this assisted in giving you a better understanding of the history of chemistry, and what could be done differently (three paragraph maximum).
Lavoisier has been considered by many scholars to be the father of chemistry. Chemists continued to discover new compounds in the 1800s. The science also began to develop a more theoretical foundation. It was in 1807, John Dalton put forth his atomic theory.
It was not until the era of the ancient Greeks that we have any record of how people explained the chemical changes they observed and used. At that time natural objects were thought to consist of only four basic elements like earth, air, fire and water.
It was in the fourth century BC, two Greek philosophers Democritus and Leucippus suggested that matter was not infinitely divisible into smaller particles but instead consists of the fundamental particles called the atoms. Chemistry took its present form in the 18th century.
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Copper is low down in the reactivity series and can be obtained from
copper oxide.
Devise a simple method to obtain a sample of copper from copper oxide in
the laboratory.
Answer:
reducing it with a reducing agent